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Travka [436]
3 years ago
13

Calculate the amount of heat (kcal) released when 50.0g of steam at 100*c hits the skin, condenses, and cools to a body temperat

ure of 37*c
Physics
2 answers:
abruzzese [7]3 years ago
7 0

conversion of steam at 100 C to water at 100 C :

m = mass of the steam = 50 g

L = latent heat of condensation of steam to water = 79.7 J/g

Heat released in conversion of steam to water is given as

Q₁ = mL

inserting the values

Q₁ = (50) (79.7)

Q₁ = 3985 cal


conversion of water at 100 C to water at 37 C :

ΔT = change in temperature = 100 - 37 = 63 C

c = specific heat of water = 1 cal/(gC)

m = mass of water = 50 g

Heat released in change of temperature is given as

Q₂ = m c ΔT

Q₂ = 50 x 1 x 63

Q₂ = 3150 cal

total heat released is given as

Q = Q₁ + Q₂

Q = 3985 + 3150

Q = 7135 cal

luda_lava [24]3 years ago
6 0
As the steam touches the skin, it undergoes a phase change and releases latent heat due to the phase change. As it reaches equilibrium, it releases sensible heat. We calculate as follows:

Q = latent heat + sensible Heat
Q = 2.26 kJ / g (50.0 g) + 50.0 g ( 4.18 J / g C) (37 C - 100 C) ( 1 kJ / 1000 J)
Q = 99.833 kJ
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3 years ago
Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left
Dahasolnce [82]

Answer:

x_c= \dfrac{5}{9}L

I=\dfrac {7}{12}\lambda_ 0 L^3

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any  distance x from point A mass density

\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x

\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

dI=\lambda x^2dx

So total moment of inertia

I=\int_{0}^{L}\lambda x^2dx

By putting the values

I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx

By integrating above we can find that

I=\dfrac {7}{12}\lambda_ 0 L^3

Now to find location of center mass

x_c = \dfrac{\int xdm}{dm}

x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}

Now by integrating the above

x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}

x_c= \dfrac{5}{9}L

So mass moment of inertia I=\dfrac {7}{12}\lambda_ 0 L^3 and location of center of mass  x_c= \dfrac{5}{9}L

8 0
3 years ago
Please. Physics is so difficult.
Softa [21]

Answer:

0.010 m

Explanation:

So the equation for a pendulum period is: y=2\pi\sqrt{\frac{L}{g}} where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:

1.99 s = 2(3.14)\sqrt{\frac{L}{9.8 m\backslash s^2}}\\

Evaluate the multiplication in front

1.99 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}

Divide both sides by 6.28

0.317 s= \sqrt{\frac{L}{9.8 m\backslash s^2}}

Square both sides

0.100 s^2= \frac{L}{9.8 m\backslash s^2}

Multiply both sides by m/s^2  (the s^2 will cancel out)

0.984 m = L

Now now let's find the length when it's two seconds

2.00 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}}

Divide both sides by 6.28

0.318 s = \sqrt{\frac{L}{9.8 m\backslash s^2}

Square both sides

0.101 s^2 = \frac{L}{9.8 m\backslash s^2}

Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)

0.994 m = L

So to find the difference you simply subtract

0.984 - 0.994 = 0.010 m

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